Let $p : \mathbb R \to \mathbb R$ be a non-constant polynomial, that is, for all $x \in \mathbb R$;$p(x) =a_0 + a_1x +... + a_nx^ n$, with $a_n \neq 0$ and $n \ge 1$. Prove that if $n$ is even then $lim_{x \to + \infty} p(x) =lim_{x \to - \infty} p(x) = + \infty$ if $a_n \gt 0 $ and = $- \infty$ if $an \lt 0$.
what was i able to do
We write
$P(x) = a_n x^n( \sum_{k=0}^{n-1} \frac{a_k}{a_nx^{n-k}}+1)$.If $n$ is even $lim_{x \to \infty} x^na_n = \infty =lim_{x \to -\infty} x^na_n $ with $a_n \gt 0$ and $lim_{x \to \infty}x^n a_n = - \infty = lim_{x\to - \infty } x^n a_n$ if $a_n \lt 0$, so the same goes for $P(x)$ If $n$ is odd $lim_{x \to \infty} x^na_n = \infty$ and $lim_{x \to -\infty}x^n a_n = -\infty$ with $a_n \gt 0$, case $a_n \lt 0$ we have $lim_{x \to \infty}x^na_n = -\infty$ and $lim_{x \to - \infty} x^n a_n = \infty$.
How it is?
Thanks.
Your question and proof is quite clumsy. So, here's my attempt at a neat argument-
Since even polynomials are even functions, it suffices to check only for $x\to \infty$.
Now, as $x^n$ grows faster than $x^k$ for any $k<n$, $$\lim_{x\to \infty} p(x)=\lim_{x\to \infty} a_nx^n$$ and hence $$\lim_{x\to \infty} p(x)=\begin{cases} +\infty &\text{ if $a_n>0$}\\ -\infty &\text{ if $a_n<0$} \end{cases}$$
If you feel that the "$x_n$ grows faster than $x^k$" part is not rigorous enough, try making it so- if you're unsuccessful, I will place an edit.